# How do you differentiate h(x)=((x^3+1)sqrtx)/x^2 using the quotient rule?

Mar 16, 2018

$h ' \left(x\right) = \frac{3 \sqrt{x} \left({x}^{3} - 1\right)}{2 {x}^{3}}$

#### Explanation:

$h \left(x\right) = \frac{\left({x}^{3} + 1\right) \sqrt{x}}{x} ^ 2$

$= \frac{{x}^{3} + 1}{x} ^ \left(\frac{3}{2}\right)$

Apply the quotient rule.

$h ' \left(x\right) = \frac{{x}^{\frac{3}{2}} \cdot \left(3 {x}^{2}\right) - \left({x}^{3} + 1\right) \cdot \left(\frac{3}{2} {x}^{\frac{1}{2}}\right)}{x} ^ 3$

$= \frac{3 {x}^{\frac{7}{2}} - \frac{3}{2} \sqrt{x} \left({x}^{3} + 1\right)}{x} ^ 3$

$= \frac{6 {x}^{\frac{7}{2}} - 3 \sqrt{x} \left({x}^{3} + 1\right)}{2 {x}^{3}}$

$= \frac{6 {x}^{\frac{7}{2}} - 3 {x}^{\frac{7}{2}} - 3 \sqrt{x}}{2 {x}^{3}}$

$= \frac{3 {x}^{\frac{7}{2}} - 3 \sqrt{x}}{2 {x}^{3}}$

$= \frac{3 \sqrt{x} \left({x}^{3} - 1\right)}{2 {x}^{3}}$

Mar 16, 2018

${h}^{'} \left(x\right) = \frac{3 \left({x}^{3} - 1\right)}{\left(2 \sqrt{x}\right) {x}^{2}} = \frac{3}{2} \cdot {x}^{\frac{1}{2}} - \frac{3}{2} \cdot {x}^{- \frac{5}{2}}$, (simplify)
OR
$h \left(x\right) = \frac{\left({x}^{3} + 1\right) \sqrt{x}}{x} ^ 2 = \frac{{x}^{3} \sqrt{x}}{x} ^ 2 + \frac{\sqrt{x}}{x} ^ 2 = {x}^{3 + \frac{1}{2} - 2} + {x}^{\frac{1}{2} - 2} = {x}^{\frac{3}{2}} + {x}^{- \frac{3}{2}} \implies {h}^{'} \left(x\right) = \frac{3}{2} \cdot {x}^{\frac{1}{2}} - \frac{3}{2} \cdot {x}^{- \frac{5}{2}}$

#### Explanation:

Here, $h \left(x\right) = \frac{\left({x}^{3} + 1\right) \sqrt{x}}{x} ^ 2$
Let,$f \left(x\right) = \left({x}^{3} + 1\right) \sqrt{x} \mathmr{and} g \left(x\right) = {x}^{2} \implies {g}^{'} \left(x\right) = 2 x$
andf^'(x)=(x^3+1)d/(dx)(sqrtx)+sqrtx*d/(dx)((x^3+1)
$\implies {f}^{'} \left(x\right) = \left({x}^{3} + 1\right) \cdot \frac{1}{2 \sqrt{x}} + \sqrt{x} \cdot 3 {x}^{2} = \frac{{x}^{3} + 1 + 2 x \cdot 3 {x}^{2}}{2 \sqrt{x}}$
$\implies {f}^{'} \left(x\right) = \frac{{x}^{3} + 1 + 6 {x}^{3}}{2 \sqrt{x}} = \frac{7 {x}^{3} + 1}{2 \sqrt{x}}$
We know that, If $h \left(x\right) = \frac{f \left(x\right)}{g \left(x\right)} , t h e n ,$
${h}^{'} \left(x\right) = \frac{g \left(x\right) \cdot {f}^{'} \left(x\right) - f \left(x\right) \cdot {g}^{'} \left(x\right)}{{\left[g \left(x\right)\right]}^{2}} = \frac{{x}^{2} \left(\frac{7 {x}^{3} + 1}{2 \sqrt{x}}\right) - \left({x}^{3} + 1\right) \sqrt{x} \cdot 2 x}{{\left({x}^{2}\right)}^{2}} = \frac{{x}^{2} \left(7 {x}^{3} + 1\right) - \left({x}^{3} + 1\right) 2 x \cdot 2 x}{\left(2 \sqrt{x}\right) \cdot {x}^{4}}$
$= \frac{7 {x}^{5} + {x}^{2} - \left({x}^{3} + 1\right) 4 {x}^{2}}{\left(2 \sqrt{x}\right) {x}^{4}} = \frac{7 {x}^{5} + {x}^{2} - 4 {x}^{5} - 4 {x}^{2}}{\left(2 \sqrt{x}\right) {x}^{4}} = \frac{3 {x}^{5} - 3 {x}^{2}}{\left(2 \sqrt{x}\right) {x}^{4}} = \frac{3 {x}^{2} \left({x}^{3} - 1\right)}{\left(2 \sqrt{x}\right) {x}^{4}} = \frac{3 \left({x}^{3} - 1\right)}{\left(2 \sqrt{x}\right) {x}^{2}}$